Question: The equation of a circle $C$ is $x^2+y^2+2x-4y+4 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2+2x) + (y^2-4y) = -4$ $(x^2+2x+1) + (y^2-4y+4) = -4 + 1 + 4$ $(x+1)^{2} + (y-2)^{2} = 1 = 1^2$ Thus, $(h, k) = (-1, 2)$ and $r = 1$.